Question

If the lines 2x-3y=5 and 3x-4y=7 are the diameters of a circle of area 154 sq. units, then obtain the equation of the circle.

Answer 1

As 2x - 3y =5 and 3x-4y=7 are the diameters of a circle
=> Their point of intersection is the center of circle
=> On solving equations we get,

2x-3y=5 ----(i) x 3
3x-4y=7 ----(ii) x 2

6x - 9y = 15 ---(iii)
6x - 8y = 14 ---(iv)

On subtracting equations (iv) from (iii) gives ,
( 6x - 9y ) - ( 6x - 8y ) = 15 - 14
=>6x - 9y - 6x + 8y = 1
=> - y = 1 => y = - 1,On putting the value of y in (i), gives
2x - 3(- 1) = 5
=> 2x + 3 = 5 => 2x = 5 - 3 => 2x = 2 => x = 1

Thus Center of a circle is ( x , y ) = ( 1 , -1 )

Now Area of circle = 154 sq. units
=> ( Pie ) ( r ^ 2 ) = 154
=> ( 22/7 ) x ( r ^2 ) = 154
=> r^2 = 154 x 7 / 22
= 14 x 7 / 2 ( on dividing 154 and 22 by 11)
= 7 x 7
=> r = 7 units
Finally, The equation of a circle having center ( 1 , - 1 ) and with radius r = 7 units is
${(x - 1)}^{2} + {(y - ( - 1))}^{2} = {7}^{2} \\ {(x - 1)}^{2} + {(y + 1))}^{2} = 49 \\ {x}^{2} - 2x + 1 + {y}^{2} + 2y + 1 = 49 \\ {x}^{2} + {y}^{2} - 2x + 2y + 2 = 49 \\ {x}^{2} + {y}^{2} - 2x + 2y + 2 - 49 = 0 \\ {x}^{2} + {y}^{2} - 2x + 2y - 47 = 0$

Hope this help.......

Answer 2

Answer:

Step-by-step explanation:

Center is the point of intersection of diagonals.

⇒2x−3y=5

⇒3x−4y=7

Solving simultaneous equation, x=1,y=−1

Center =(1,−1)

For radius r ( let )

⇒r=  

π

154

​  

 

​  

  ⇒  

22

154

​  

×7

​  

=7

Equation -

⇒(x−1)  

2

+(y+1)  

2

=7  

2

 

⇒x  

2

+1−2x+y  

2

+1+2y=49

⇒x  

2

+y  

2

−2x+2y=49−2

⇒x  

2

+y  

2

−2x+2y=47

Hence, the answer is x  

2

+y  

2

−2x+2y=47.